# Thread: Find the Galois group?

1. ## Find the Galois group?

I have the polynomail f(x)=x^5 +2x+2 over rationals. Let M be teh splitting field of this over rationals.

Then I want to show that the galois group is S_5.

My argument is as follows:

f(x) is irreducible by Eisenstein with p=2. and the galois group is isomorphic to a subgroup of S_5, let us call it H.

we have that the order of H is divisible by 5, then by using Cauchy's theorem we have that there is an element of order 5 in H, so it must be a 5-cycle. So H contains a five cycle.

Now I can show that f has only one real root?

Then what will be the argument be? I want to conclude something like that H contains a transposition, so H must be S_5.

Any help, any hints? Thanks.

2. Originally Posted by ZetaX
I have the polynomail f(x)=x^5 +2x+2 over rationals. Let M be teh splitting field of this over rationals.

Then I want to show that the galois group is S_5.

My argument is as follows:

f(x) is irreducible by Eisenstein with p=2. and the galois group is isomorphic to a subgroup of S_5, let us call it H.

we have that the order of H is divisible by 5, then by using Cauchy's theorem we have that there is an element of order 5 in H, so it must be a 5-cycle. So H contains a five cycle.

Now I can show that f has only one real root?

Then what will be the argument be? I want to conclude something like that H contains a transposition, so H must be S_5.

Any help, any hints? Thanks.
Modulo 29 this polynomial factors into irreducibles as $(x^2+4x+7)(x^3+25x^2+9x+21)$.
By Dedekind's theorem its Galois group has a permuation that is a product of a 2-cycle and a 3-cycle.
The only transitive subgroup of $S_5$ with this property is $S_5$ itself.

3. Thanks for your help, but what abiut the way I started with, doesnot it work in this case?

How can i show that the only transitive subgroup of S_5 contaning a 3 and 2 cycle is S_5?